Jambmaths question:
In the diagram above, O is the centre of the circle, POM is a diameter and $\angle MNQ={{42}^{\text{o}}}$. Calculate $\angle QMP$
Option A:
42o
Option B:
48o
Option C:
132o
Option D:
138o
Jamb Maths Solution:
$\begin{align} & \angle MNQ={{42}^{\text{o}}} \\ & \angle MPQ=\angle MNQ={{42}^{o}}\text{ }\!\!\{\!\!\text{ }\angle \text{s in the segment }\!\!\}\!\!\text{ } \\ & \angle MQP={{90}^{o}}\text{ }\!\!\{\!\!\text{ }\angle \text{s in a semicircle }\!\!\}\!\!\text{ } \\ & \angle MPQ+\angle MQP+QMP={{180}^{o}}\text{ }\!\!\{\!\!\text{ Sum of }\angle \text{s in a }\vartriangle \text{ }\!\!\}\!\!\text{ } \\ & {{42}^{o}}+{{90}^{o}}+\angle QMP={{180}^{o}} \\ & \angle QMP={{48}^{o}} \\\end{align}$
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